Question 911108: Find (x,y) position of centroid of the area under the curve y=1+x+x^2 from x=0,x=2.?
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! note that the graph of y=1+x+x^2 is completely above the x axis so we can proceed with calculating the area under the curve. First integrate,
y = 1 +x +x^2
integral is x +x^2/2 +x^3/3
we evaluate the integral for x=0 and x=2
the integral is 0 for x=0 and 2+4/2+8/3 = 8/2 +8/3 = 24/6 + 16/6 = 40/6 = 20/3
therefore the area(A) under the curve for x=0 and x=2 is 20/3
to find the (x,y) position of the centroid, we use the following two equations
x = (1/A)*integral of x*(1+x+x^2)dx
y = (1/A)*integral of ((1+x+x^2)^2)/2 dx
a) x = (3/20) * (x^2/2 + x^3/3 + x^4/4), we evaluate the polynomial in parenthesis first for x = 0 and x = 2. The polynomial is 0 for x = 0, then
x = (3/20) * (2+(8/3)+4) = 3/20 * 26/3 = 26/20 = 13/10 = 1.3
b) y = (3/20) * ((x^4 +2x^3 +3x^2 +2x + 1)/2), we evaluate the polynomial in parenthesis first for x = 0 and x = 2. The polynomial is 0 for x = 0, then
y = (3/20) * ((16+16+12+6+1)/2)) = (3/20) * (51/2) = 3.825
therefore,
(x,y) = (1.3, 3.825)
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